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The Six-fold Basis of Reality: The Proton Revisited
Preprint · June 2023
DOI: 10.13140/RG.2.2.32939.41762
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Ian Beardsley
University of Oregon
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! of 1 66
The Six-fold Basis of Reality: The Proton Revisited
By
Ian Beardsley
Copyright © 2023 by Ian Beardsley!
! of 2 66
Contents
Introduction………………………………………………………3
1.0 Inertia………………………………………………………….4
2.0 The Constant k……………………………………………..8
3.0 The Proton Radius…………………………………………12
4.0 Discussion…………………………………………………….15
5.0 Solar Eclipsing Moon……………………………….22
6.0 The Proton Charge…………………………………..27
7.0 The Solar Magnetic Field………………………….29
8.0 Conclusion……………………………………………..41
9.0 The Proton Revisited……………………………….43
Appendix 1………………………………………………………….63
Appendix 2………………………………………………………….64
! of 3 66
Introduction
Here I hypothesize an equation of state for the periodic table of the elements from Avogadro’s
number, and determine from that a constant k based on the idea that there is a constant for the
intermediary scale between protons and stars using the Chandrasekhar limit for white dwarf
stars. After constructing a six-fold theory for inertia, or mass, based on the idea that a proton is
a three dimensional cross-section of four dimensional hypersphere, I successfully predict the
radius of a proton. In the course of this we find the second is a Natural unit of time both in terms
of the proton, and gravitational constant G and Planck’s constant h, and in terms of the Earth/
Moon orbital mechanics. I show Avogadro’s number is a natural unit as well, both for the proton
and atom, the equation of state, or constant k, and for the macroscale of Moon/Earth/Sun. We
also see that this is a consequence of six-fold symmetry based on carbon the core element of life,
and hydrocarbons, the skeleton’s or biological life chemistry. We find as well we can determine
the charge of a proton with our constant k in terms of the six-fold, as well as the function of the
solar magnetic field at Earth orbit.
I really think Nature employs 6-fold symmetry because I think it is the most dynamic being the
product of the two smallest prime numbers 2 and 3. Those are the smallest factors down to
which anything can be reduced, so that is why I think 6-fold is the basis of Nature, that the basis
has to be reduced to the smallest factors. As such we have 2x3=6, 3x3=9, 2x9=18 and 3x6=18.
The periodic table of elements is periodic over 18 groups, which means when you count to 18 the
elements start over making groups where they have similar properties falling into the same
groups. These groups are determined by electron configurations, for instance carbon is in group
4 which means since it wants to have noble gas electron configuration, it gains 4 electrons to be
the same as the noble gas in group 18 of the same period. 18-14=4 meaning it ionizes as C4- so it
has 4 electrons to combine with 4 hydrogen atoms that are each H+ or so we can have two H
atoms that can combine with one oxygen atom which is O2- because oxygen is in group 16
meaning 18-16=2, and so forth making life possible.
Here in this version of the paper we revisit the proton and go into an explanation of the Van Der
Waal Radius from which our equation for the radius of a proton was derived. We also explain
how the Chandrasekhar limit was derived, but don’t derive it and leave that to the textbooks in
Astrophysics that all derive it in their chapters on stellar dynamics. The Chandrasekhar limit is
used here for the derivation of our intermediate mass, our constant k, which determines our
derivation of the unit of a second of time, and the charge of a proton, as well as the radius of a
proton. We find in this second look at the proton we can hone our equation for the radius of a
proton for more accuracy and what might be a more dynamic expression than the original
equation. We then look at the geometry that follows from it.!
! of 4 66
1.0 Inertia
We now need to show that the fundamental particles that build reality are based on sixfold
symmetry for their mass, size, and charge and we want to do it in terms of gravity on the
macroscale, thus it has to use G the universal constant of gravitation and Planck’s constant h,
that quantizes energy on the microscale. I find I can do this as such:
The natural constants measure the properties of space and time. We can list some of them here:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Gravity is a property of space measured by the universal constant of gravity, G:
Equation 1.0
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 1.2
Which describes mass per meter over time, which is:
Equation 1.3
It must be adjusted by the fine structure constant . It is my guess the factor should be
which is 18,769.:
Equation 1.4
Because the fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model times the
speed of light squared, that is it represents the ground state. It is
m
P
: 1.67262 × 10
27
kg
r
p
: 0.833 × 10
15
m
G:6.67408 × 10
11
N
m
2
kg
2
c:299,792, 459m /s
α : 1/137
q
p
= q
e
= 1.6022E 19coulombs
k
e
= 8.988E9
Nm
2
C
2
G = 6.674E 11N
m
2
s
2
= 6.674E 11
m
3
s
2
kg
h
Gc
=
kg s
m
1.82E 16
kg s
m
α
1
α
2
(1.82E 16kg s/m)(18,769) = 3.416E 12kg s/m
! of 5 66
Equation 1.5
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 1.6
We take the square root to get meters:
Equation 1.7
We multiply that with the value we have in equation 1.4:
Equation 1.8
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 1.9
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 1.10
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
Equation 1.11
Equation 1.12
It seems the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that (See Appendix 1):
Equation 1.13
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds). The earth day changes very little, by
very small amounts over millions of years. The solar system has evolved towards this since the
explosion of life called the Cambrian, and will slowly decay away from it. But we need to derive
the second in terms of something else. For now we have the mass of a proton as:
α
2
=
U
e
m
e
c
2
S
p
S
p
= 4π r
2
p
= 8.72E 30m
2
S
p
= 2.953E 15m
(2.953E 15m)(3.416E 12
kg s
m
) = 1.009E 26kg s
m
p
m
h
=
1E 26
6secon ds
= 1.67E 27kg
m
c
=
1E 26kg s
1secon d
= 1E 26kg = 6protons = 6m
p
m
p
=
1E 26kg s
6secon ds
m
c
=
1E 26kg s
1secon d
KE
moon
KE
earth
(EarthDay) 1second
! of 6 66
Equation 1.14
This way of looking at things is to say matter is that which has inertia. This means it resists
change in position with a force applied to it. The more of it, the more it resists a force. We
understand this from experience, but what is matter that it has inertia? In this analogy we are
suggesting a proton is a three dimensional bubble embedded in a two dimensional plane. As
such there has to be a normal vector holding the higher dimensional sphere in a lower
dimensional space. (See Fig. 1) Thus if we apply a force to to the cross-section of the sphere in
the plane there should be a force countering it proportional to the normal holding it in a lower
dimensional universe. It is actually a 4-dimensional hypersphere whose cross-section is a
sphere. This counter force would be experienced as inertia. Our reasoning above in one equation
is:
Equation 1.15
That is 1 second gives carbon. We find six seconds gives 1 proton is hydrogen:
Equation 1.16
m
p
=
3r
p
18α
2
4πh
Gc
1
α
2
m
p
h4π r
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
Fig. 1
! of 7 66
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements. This is another indication that a second is natural; Carbon and hydrogen are the
upper and lower limits for integer seconds.
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving (Program in Appendix 2):
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
So again, a very interesting thing here is looking at the values generated by the program, the
smallest integer value 1 second produces 6 protons (carbon) and the largest integer value 6
seconds produces one proton (hydrogen). Beyond six seconds you have fractional protons, and
the rest of the elements heavier than carbon are formed by fractional seconds. These are the
hydrocarbons the backbones of biological chemistry. So we have the six-fold symmetry in the
chemical skeletons of life.This is to say that Carbon equals one second produces the radius of a
proton.
Equation 1.17
Where the above is multiplied by one second to make the units work. Where t is exactly:
Equation 1.18 !
1
α
2
m
p
h4π r
2
p
Gc
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
! of 8 66
2.0 The Constant k
Warren Giordano wrote in his paper The Fine Structure Constant And The Gravitational
Constant: Keys To The Substance Of The Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where
’ is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
I found I could eliminate the and at the same time get the six of the six-fold symmetry with
which I was working by considering Avogadro’s number .
I suggested there exists some k that serves as a constant that describes both the microcosmos
and macrocosmos from the proton, to the atoms, to planetary orbits. It is such that the square
root of it times the earth orbital velocity is 6, because we are guessing we are dealing with six-
fold symmetry as the basis of Nature. That is
Eq. 2.1
We have that k is
Eq. 2.2
This follows from what Warren Giordano noticed that
Eq. 2.3
Without the right units. I noticed since Avogadro’s number is that I
could introduce an equation of state for the periodic table of the elements:
Eq. 2.4
Eq 2.5
Let us say we were to consider Any Element say carbon . Then in general
We have
and
h
1 + α
α
10
23
10
23
6.02E 23atoms
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
h(1 + α) 10
23
= G
6.02 × 10
23
6 × 10
23
= 1
gram
atom
h
(1 + α)
G
N
A
H = 6.0003
kg
2
s
m
𝔼
h
(1 + α)
G
N
A
𝔼 = 6.0003kg
2
s
m
=
6gram s
6protons
N
A
=
6(6E 23protons)
6gram s
! of 9 66
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And
Therefore we always have:
is a variable, the number of protons in multiplied by Avogadro’s number.
Put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Equation. 2.6
While we have masses characteristic of the microcosmos like protons, and masses characteristic
of the macrocosmos, like the upper limit for a star to become a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 solar masses — More mass than that and she will collapse —
we do not have a characteristic mass of the intermediary world where we exist, a truck weighs
several tons and tennis ball maybe around a hundred grams. To find that mass let us take the
geometric mean between the mass of a proton and the mass of 1.44 solar masses. We could take
the average, or the harmonic mean, but the geometric mean is the squaring of the proportions, it
is the side of a square with the area equal to the area of the rectangle with these proportions as
its sides. We have:
Equation. 2.7
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Equation 2.8
N
A
= 6E 23
h
(1 + α)
G
N
A
= 6.0003kg
2
s
m
𝔼
N
A
=
Z 6E 23protons
Z gram s
𝔼 =
Z gram s
Z protons
N
A
𝔼 = 6E 23
N
A
𝔼
h
(1 + α)
G
N
A
𝔼 v
e
= 422.787kg
M
= 1.98847E30kg
m
p
= 1.67262E 27kg
m
i
= (2.8634E 30)(1.67262 E 27) = 69.205kg
! of 10 66
All we really need to do now is divide equation 2.6 by equation 2.8 and we get an even number
that is the six of our six-fold symmetry.
Equation. 2.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Eq. 2.10
Where k is a constant, given
Eq. 2.11
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the electron degeneracy pressure and collapse. The non-
relativistic equation is:
Let us approximate 0.77 with 3/4. Since we have our constant
Eq 2.12
Eq. 2.13
Then
Eq. 2.14
Since our constant k in terms the Chandrasekhar limit is
1
m
i
h
(1 + α)
G
N
A
𝔼 v
e
= 6.1092 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
773.5
s
m
M 0.77
c
3
3
G
3
N
m
4
p
= 1.41
k =
1
m
2
i
h
1 + α
G
N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
3
G
3
m
2
p
1/2
= h /2π
! of 11 66
Eq. 2.15
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared, that is it represents the ground state. It is
Since
We are suggesting the earth orbit is the ground state for our planetary system. We suggest it
holds for any planetary system because k as we will see is a natural constant that solves many
physical problems on many levels, not just planetary systems but atomic systems and the
particles that make them up.
Let us now recall equations 1.15 and 1.16
While we have considered them to be proton-seconds because they are a mass divided by the
mass of a proton, we can consider these two masses to cancel and say they are equal to 1 second
and six seconds respectively. We have that carbon, which is to evaluate them at one second, is
the radius of a proton:
Eq. 2.16
This gives the radius of a proton is:
Eq 2.17.
Where . The experimental value of the proton radius is 0.833fm+/-0.014fm!
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
α
2
=
U
e
m
e
c
2
k v
e
= 6
1
α
2
m
p
h4π r
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
1
α
2
m
p
h4π r
2
p
Gc
= 6secon ds
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
= 1secon d
! of 12 66
3.0 The Proton Radius
Thus we have the radius of a proton is given by carbon by evaluating at one second:
But to get that we have to multiply by one second and we need one second in terms of the atom
for a theory of the proton. I find we can do that…
Substitute for to get
We have now introduced the radius of a hydrogen atom . It seems we have to
divide by two which I think is because we are looking at packing of atoms. This radius of the
hydrogen atom is the Van Der Waals radius, which is the closest distance between two hydrogen
atoms noncovalently bound. It is 120 pm. Divide that by ck where 1/k is our constant
And we find
We have our equation for the radius of a proton
We only need to multiply it by to have the right units, and we get!
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
t
6
=
1
α
2
m
p
h4π r
2
p
Gc
t
6
=
r
p
α
2
m
p
h4π
Gc
R
H
/2
r
p
t =
R
H
2α
2
m
p
h4π
Gc
R
H
= 1.2E 10m
R
H
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12seconds
3 2
16
h
Gπ
1
α
2
m
2
p
R
H
N
A
𝔼
= 1.12seconds
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
t /ck = 1secon d
! of 13 66
Equation 3.1.
Then suggest we picked up 9/8 in approximations which is close to one anyway so we write
Equation 3.2.
We form constants:
And we have the Equation:
Equation 3.3
We can say that Avogadro’s number is not an arbitrary number because it is such that there are
twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add
to this section the equation of the mass of a proton.
If our equation is right and we put it into natural units then the product should be close to
one:
Let us start with the units with which we are working:
r
p
=
9
8
2
1
m
p
hc
4π
3
G
R
H
N
A
𝔼
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
= 8.26935E 16m 0.827f m
r
p
=
1
kg
(
kg
m
2
s
)
(
m
s
)
(
kg
s
2
m
3
)
R
H
N
A
=
1
kg
(
kg
)
m
k =
hc
2π
3
G
= 6.93E 9kg
R
H
N
A
𝔼
= 1.99E 34m
r
p
m
p
= k
R
H
N
A
𝔼
m
p
=
1
r
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
m
p
G =
m
3
kg s
2
h = kg
m
2
s
! of 14 66
And convert these to proton-masses and proton-radii:
Now we find k in these units:
Thus we have:
!
c = m /s
G = 6.67408E 11
m
3
kg s
2
1.67262E 27kg(0.833E 15m)
3
s = 193,131, 756
h = 6.62607E 34kg
m
2
s
s
(0.833E 15)
2
(1.67262E 27kg)
= 5.71E 23
c =
(299,792, 459m /s)(1sec)
(0.833E 15m)
= 3.6E 23
R
H
=
1.2E 10m
0.833E 15m
= 144,058
k =
hc
2π
3
G
= 6.93E 9kg
k =
(5.71E 23)(3.6E 23)
2π
3
(193131756)
= 4E18proton masses
r
p
m
p
= k
R
H
N
A
𝔼
r
p
m
p
=
(4E18)(144058)
(6E 23)
=
5.76E 23
(6E 23)
= 0.96 1
! of 15 66
4.0 Discussion
The crux is that
Is calculated with perihelion and at aphelion giving a near
perfect result for one second. And, . And,
. We have
If multiplied by one second, which gives
Where t is
Giving
Using
Where ck is m/s cancels with s/m, and k is
1
6α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
KE
moon
KE
earth
(EarthDay) = 1.08seconds
KE
moon
KE
earth
v
e
= 30,290m /s
v
m
= 966m /s
Si xDays = 6(24)(60)(60) = 518400s
6m
p
= carbon
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
t =
R
H
2α
2
m
p
h4π
Gc
r
p
= 8.26935E 16m 0.827f m
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12seconds
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
! of 16 66
But as well we can get our one second from
We notice this is centered around the structure at the basis of life, the hydrocarbons
The k is derived from coupling these two :
Which holds for any element:
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And
KE
moon
KE
earth
(EarthDay) = 1.08seconds
1
α
2
m
p
h4π r
2
p
Gc
= 6proton secon ds = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton 6secon ds = hydrogen(H )
N
A
𝔼
h
(1 + α)
G
N
A
𝔼 = 6.0003kg
2
s
m
=
6gram s
6protons
N
A
=
6(6E 23protons)
6gram s
N
A
= 6E 23
h
(1 + α)
G
N
A
= 6.0003kg
2
s
m
𝔼
N
A
=
Z 6E 23protons
Z gram s
! of 17 66
Therefore we always have:
is a variable, the number of protons in multiplied by Avogadro’s number. By using the
Chandrasekhar limit we define an intermediate mass from the mass of the Sun
We multiply this by 1.44 (1..44 solar masses is the Chandrasekhar limit) to get 2.8634E30kg.
The mass of a proton is . We have the intermediary mass is:
More explicitly using the Chandrasekhar limit:
Intermediary mass:
The constant k given explicitly:
The k is such that
And k using the approximation
We show the equation for one second in terms of the Natural constants on the next
page…!
𝔼 =
Z gram s
Z protons
N
A
𝔼 = 6E 23
N
A
𝔼
M
= 1.98847E30kg
m
p
= 1.67262E 27kg
m
i
= (2.8634E 30)(1.67262 E 27) = 69.205kg
M 0.77
c
3
3
G
3
N
m
4
p
= 1.41
m
i
=
3
2
c
3
3
G
3
m
2
p
1/2
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
k v
e
= 6
(1 + α) 1
k =
4
3
m
p
8π
3
G
c
3
h
N
A
𝔼
! of 18 66
!
! of 19 66
We discuss what interests me most on the next page, the two different equations for one
second that connect the macrocosmos (the Earth, Moon, and Sun) to the microcosmos,
the constants of the elements.!
! of 20 66
If we can predict the radius of a proton with
Equation 4.1.
Where is given by in terms of the Moon, Earth, and Sun:
Equation 4.2.
Or in terms of the Natural Physical Constants Using equations:
Equation 4.3.
Then to understand why the solar system is described in in terms of elemental states of
periodic table that we have formulated and in that this predicts the solar system as
something that can be described in a quantum mechanical manner, we equate the latter
two:
Eq. 4.4.
Eq. 4.5.
Which is
Eq 4.6
This seems to say something interesting to me. A mole does not just pertain to atoms
per gram of an element, or molecules per sample, but perhaps speaks of a number of
Moons, Planets, Stars, or even galaxies. Let’s analyze to try and determine how that
might work, we break equation 4.5 and 4.6 down into its factors (Next page):!
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
KE
moon
KE
earth
(Ear thDay) = 1.08seconds
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12seconds
KE
moon
KE
earth
(Ear thDay) =
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
N
A
𝔼 =
R
H
16πα
2
m
2
p
3 2
h
G
KE
earth
KE
moon
1
Ear thDay
6.02E 23 =
R
H
16πα
2
m
2
p
3 2
h
G
KE
earth
KE
moon
1
Ear thDay
! of 21 66
We see that we might be speaking of 4.286E20 meters per kg mass perhaps in a Nebula
where stars form:
Equation 4.7.
We see the scaling is given by h/G by a factor of which is like comparing the small
(h) to the large (G):
Equation 4.8.
And finally this is given by a factor of the Earth to the Moon:
Equation 4.9.
There is also a fraction of:
Equation 4.10.
And, there is a frequency given by the Earth day, the time it takes to rotate:
We find something interesting. If these factors describe the Solar System the way
Avogadro’s number describes elements and their compounds, then it would make sense
because their product is Avogadro’s number and as such it applies not just to elements
and compounds but to planets and stars and the interstellar medium. And we see here
that
Describes the KE of Earth and Earth rotation period in terms of Avogadro’s number, the
mass of a proton, the radius of a hydrogen atom, and the mass of a proton, the latter two
being the primary constituents of the Universe, gas giant planets, and the interstellar
medium.!
R
h
16πα
2
m
p
=
(18769)1.2E 10m
16π(1.672E 27kg)
= 2.67875E19
m
kg
1/m
p
h
Gm
p
=
(6.626E 34)
(6.674E 11)(1.672E 27)
= 5,938kg
s
m
KE
earth
KE
moon
=
(5.972E 24kg)(30,290)
2
(7.34767E22kg)(966m /s)
2
= 79,912 . 5
3 2 = 4.2426
1
Earth Day
=
1
86,400
= 1.1574E 5
c ycles
s
6.02E 23 =
R
H
16πα
2
m
2
p
3 2
h
G
KE
earth
KE
moon
1
Ear thDay
! of 22 66
5.0 Solar Eclipsing Moon
Essentially that the Moon perfectly eclipses the Sun as seen from the Earth means while
it is 400 times smaller than the Sun it is 400 times further from the Sun than it is from
the Earth. This determines its orbital velocity and mass, as well as that of the Earth and
the mass of the Sun. The orbital velocities of the Moon and the Earth are given by:
Equation 5.1. and
Equation 5.2.
The Moon perfectly eclipses the Sun because
Equation 5.3.
Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius,
and is the lunar radius. This gives:
Equation 5.4.
We have more explicitly:
Equation 5.5.
The rotational period of the moon is and the
orbital period of the Earth is is 27.3 which gives
v
e
=
GM
r
e
v
m
=
GM
e
r
m
v
e
v
m
=
M
M
e
r
m
r
e
M
M
e
r
m
r
e
=
1.989E30kg
5.972E 24
1.74E6m
6.957E8m
= 28.86
r
e
r
m
R
R
m
r
e
r
m
R
R
m
v
e
v
m
=
M
M
e
R
m
R
= 28.6
v
e
v
m
=
R
m
R
M
M
e
r
e
r
m
R
m
R
M
M
e
r
e
r
m
=
1.74E6m
6.957E8
1.989E30kg
5.972E 24kg
1.496E11m
3.84748E8m
= 28.46
T
m
= 27.3days = 2,358, 720seconds
Γ
e
= 24hours = 86,400seconds
! of 23 66
Equation 5.6.
Thus while second is given by:
The Lunar month is given by
Equation 5.7.
We also found in terms of the proton, the second is given by
Which gives the radius of the proton as
Where is one second approximated by
Which gives the radius of a proton is
Which we derived using our constant k:
T
m
Γ
e
=
R
m
R
M
M
e
r
e
r
m
KE
moon
KE
earth
(Ear thDay) = 1.08seconds
T
m
=
R
m
R
M
M
e
r
e
r
m
Γ
e
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12seconds
r
p
=
9
8
2
1
m
p
hc
4π
3
G
R
H
N
A
𝔼
k =
4
3
m
p
8π
3
G
c
3
h
N
A
𝔼
! of 24 66
The Earth day is then:
Equation 5.8.
But since we also have
Then we have:
Equation 5.9
Where
Which gives
Equation 5.10
Equation 5.11
!
Γ
e
=
R
R
m
M
e
M
r
m
r
e
T
m
KE
moon
KE
earth
(Ear thDay) = 1.08seconds
KE
moon
KE
earth
R
R
m
M
e
M
r
m
r
e
T
m
= 1.08s
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12seconds
KE
moon
KE
earth
R
R
m
M
e
M
r
m
r
e
T
m
=
1
6α
2
m
P
h4πr
2
p
Gc
KE
moon
KE
earth
R
R
m
M
e
M
r
m
r
e
T
m
=
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
! of 25 66
From equation 5.8 for the Earth day
Checking our work we see from equation 5.7 for the Earth month
One of the things that we have not done yet is equate the expression on the right of 5.10
with that on 5.11
Which is Avogadro’s number in terms of the size of a hydrogen atom, the mass of a
proton, the radius of a proton, and h and G:
Equation 5.12
Let see just what kind of prediction 5.10 brings
Γ
e
=
R
R
m
M
e
M
r
m
r
e
T
m
T
m
=
R
m
R
M
M
e
r
e
r
m
Γ
e
Γ
e
=
R
R
m
M
e
M
r
m
r
e
R
m
R
M
M
e
r
e
r
m
Γ
e
1 = 1
1
6α
2
m
P
h4πr
2
p
Gc
=
R
H
16πα
2
m
2
p
3 2
N
A
𝔼
h
G
N
A
𝔼 =
9 2R
H
8πr
p
m
p
hc
4πG
KE
moon
KE
earth
R
R
m
M
e
M
r
m
r
e
T
m
= 1.08s
KE
moon
KE
earth
R
R
m
M
e
M
r
m
r
e
T
m
=
1
6α
2
m
P
h4πr
2
p
Gc
! of 26 66
=28.46
=
We convert that to days,
2.28568E6/{(60s)(60m)(24d)}=31.74556 days
The lunar month with respect to the Sun (sidereal month) is 27.3 days. And with respect
to the Earth (synodic month) is 29.53059 days. So we are in the right area. We have
suggested the second is a Natural unit, but we might say the month is as well. In a sense
we already knew this because we have always known the Moon perfectly eclipses the
Sun, and here we have shown that is determined by the mass of the Earth, the radius of
the Sun, its mass, and these determine the Moon’s orbital distance, size, and mass, as
well the Earth’s mass. The Moon also has a function; it allows for life on Earth because
its orbit holds the Earth’s inclination to its orbit which allows for the seasons.
T
m
=
1
6α
2
m
P
h4πr
2
p
Gc
KE
earth
KE
moon
R
m
R
M
M
e
r
e
r
m
KE
earth
KE
moon
=
(5.972E 24kg)(30,290)
2
(7.34767E22kg)(966m /s)
2
= 79,912 . 5
1
6α
2
m
p
h4πr
2
p
Gc
= 1.004996352seconds
R
m
R
M
M
e
r
e
r
m
=
1.74E6m
6.957E8
1.989E30m
5.972E 24m
1.496E11m
3.84748E8m
T
m
= (1.005s)(79,912 . 5)(28.46) = 2.28568E6seconds
! of 27 66
6.0 The Proton Charge
I construct the electric field as such (Fig. 2): is one component of . Events in
are through time with components and where c is the speed of light. The
electrons and the protons and in the presence of one another cross into giving
themselves acceleration in the and directions.
We have
We suggest
Equation 6.1
The position of changes in the space of and has travelled to a place in
time, as well. Since ct=meters there is a g such that . Thus since
x
3
(x, y, z)
3
(x, y, z)
ct
1
ct
2
q
e
q
p
ct
1
ct
2
x
x
i
j
k
0 ct
1
0
0 0 ct
2
= (ct
1
ct
2
)
i
··
x (ct
1
ct
2
)
i
x
3
(x, y, z)
t (0,ct
1
, ct
2
)
··
x = g(ct
1
ct
2
)
i
g m =
m
s
2
Fig. 2
! of 28 66
and that is . Thus g is frequency squared ( ) and we suggest it is
derived from the separation between the charges and . That there is some velocity v such
that . We call upon our equation for 1 second:
Letting v= from our equation for k
And we have
Equation 6.2
We get
Equation 6.3 !
g =
1
s
2
=
1
t
2
1
s
2
m
s
s =
m
s
2
f
2
q
e
q
p
f =
(
v
x
)
r
p
m
p
h 4π
Gc
= 1secon d
(
v
x
)
2
(c)
r
p
m
p
h 4π
Gc
=
k
e
m
q
2
x
2
v
2
(c)
r
p
m
p
h 4π
Gc
= k
e
q
2
m
α
2
6
1
k
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼 =
1
773.5
s
m
1
k
= 773.5
m
s
m = m
p
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
h 4π r
2
p
Gc
=
(6.626E 34)(4π)(0.833E 15)
2
(6.674E 11)(299,792, 459)
= 5.37E 31
q = (0.033)
1
36(18769
2
)
773.5
2
(5.37E 31) = 9.1435E 19C
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71protons 6protons
! of 29 66
7.0 The Solar Magnetic Field
We model the formation of the solar system from a slowly rotating gas cloud, a nebula of
gaseous molecules, that collapses into a flat disc with a protostar at its center. The star turns on
and blows lighter elements far away, like hydrogen and helium, from which form the gas giants,
like Jupiter and Saturn, and the heavier elements stay closer in, like iron and silicates, from
which form the terrestrial planets like Venus, Earth, and Mars form. There are basically three
factors that determine its structure, the inward gravity, the pressure gradient outward which
balances with the inward gravity, and the outward inertial forces from the planets’ orbits. The
flattened rotating disc is broken up into rings each that has a mass spread out over it from which
the planets form. We estimate the ring associated with the Earth, had in its lower limit 230 earth
masses spread over it for the Earth to form. We further estimate that the Venus ring had a mass
spread over it of 230 Venus masses for Venus to form, and the Mars ring similarly had 230 Mars
masses spread out over it for Mars to form. The asteroid belt had about 200 of it masses, and the
Jovian planets 5, 8, 15, and 20 masses of each respectively. For Mercury it requires a factor of
about 350 because it is mostly iron condensations with incomplete silicon condensations.
Plotting these logarithmically we get the exponent of r, the distance of a planet from the sun is
-1.5 so that the density distribution of the protoplanetary disc is:
Giving a mass
With pressure gradients playing the key role in the formation of solar system, less attention is
payed to the magnetic field of the Sun. However, in the older literature, one of the pioneer’s of
this aspect found something very interesting concerning it. He was Alfven (1942). At the time
people were suggesting instead of the solar system forming from a rotating nebula, rather the
sun came into existence not at the same time at the center of the disc, but rather passed through
clouds and captured material after already existing. He figured for the captured material its
inward component v, and density , at a distance r from the sun, had to conserve mass, which
required:
He figured as the velocities of the atoms got closer to the sun, were moving then faster, collisions
would increase, and so temperature would go up, ionizing the atoms and therefore ionized, the
magnetic field becomes important. He considered for simplicity the solar magnetic field was
generated by a dipole moment , a vector quantity, and that a particle moving in the plane of
that vector with mass m and charge q, would have all of both the gravitational and magnetic
forces in that plane, so the problem becomes two-dimensional and required only the and , of
polar coordinates. The differential equations of its motion would be:
σ (r ) = σ
0
r
3/2
σ
0
= 3300
M =
2π
0
r
h
r
s
σ (r )r dr d θ
ρ
dM
dt
= 4π r
2
ρv
μ
θ
r
! of 30 66
Equation 7.1
And,
Equation 7.2
We can integrate equation 7.2 with the boundary condition that the angular momentum of the
particle is zero at large distances from the sun to get (Appendix 2):
Equation 7.3
And substitute it into equation equation 7.1 for to get
Equation 7.4
Which we can write
Equation 7.5
We then integrate this with respect to r with the boundary condition that at large r and get
Equation 7.6
He then notices there is another value for which . It is
Equation 7.7
This is interesting because it means the particle can never approach the Sun closer than this
value, and it depends only on the value q/m, the charge to mass ratio of a particle. He took this
as hydrogen because ionized it is a proton, for which q/m is well defined. He estimated what the
magnetic field of the Sun could have been in this earlier stage of its life, and adjusted for the fact
that hydrogen doesn’t ionize until it reaches a velocity of 5E4 m/s and found that was the
region occupied by the major planets which are Jupiter and Saturn mostly made of hydrogen
and helium.
Certainly today we don’t see the planets as having formed from material gathered by the Sun in
its journey, but rather think the Sun and planets formed at the same time from a cloud that
collapsed into a rotating flat disc. And indeed, there may be stars in the galaxy that pass through
clouds and gather material, and indeed Alfven’s equations would hold preventing ionizing
clouds from falling into their star. But we can also apply his equation to our Sun today, for which
we know a great deal about its magnetic field, which also happens to be an important thing to
study and for which we have satellites in the Lagrange points, where the Earth’s gravity cancels
m
··
r =
GM
m
r
2
+
qμ
·
θ
r
2
+ mr
·
θ
2
m
r
d(r
2
·
θ )
dt
=
qμ
·
r
r
3
mr
2
·
θ =
qμdr
r
2
=
qμ
r
·
θ
m
··
r =
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r
d
·
r
dr
=
GM
r
2
+
2q
2
μ
2
m
2
r
5
·
r = 0
·
r
2
=
2GM
r
q
2
μ
2
m
2
r
4
·
r = 0
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
r
c
! of 31 66
with that of the Sun, where the orbits are very stable, so we can understand the solar magnetic
field. It is a complex field, that interacts with the Earth’s magnetosphere, and we need to predict
solar maximums, so we have warning as to whether there will be a magnetic storm that will
knock out our electrical grid and internet, ahead of time.
During solar minimum the solar magnetic field has closed lines, that flow out one pole and into
the other. The dipole field of the sun is about 50 Gauss. There are 10,000 Gauss in a Tesla, so
that is 5E-3 Tesla. That is the magnetic field strength where the field goes into the poles. The
total magnetic field of the Sun at the Earth, is all the components taken together, which are
, , and . The important component is , because it runs north-south, so it is
perpendicular to the ecliptic, the path traced out by the sun due to the earth’s orbit. It is the
component that interacts with the Earth magnetosphere, and when it points southward, it will
connect with the Earth’s magnetosphere which points northward so the solar poles flow into the
Earth poles and the Earth field then gets disrupted allowing particles from the solar wind to rain
down along Earth magnetic field lines causing the Aurora. The solar magnetic field doesn’t
always stay around the Sun itself, but the solar wind carries it through the solar system until it
collides with the interstellar medium reaching the heliopause. Thus the Sun creates the
Interplanetary Magnetic Field (IMF) which has a spiral shape because the Sun rotates once
about every 25 days. But the upshot is that at Earth we have
Moderate Magnetic Field: 10 nT
Strong Magnetic Field: 20 nT
Very Strong Magnetic Field: 30 nT
For our purposes we want to return to equation 55:
B
t
B
x
B
y
B
z
B
z
! of 32 66
And ask just what is , because in the time that Alfven was working we worked with magnetic
fields differently, aside from his equation uses a trick, which we still use today, and that is to
consider the magnetic field a dipole. To consider it like this is to say there are two monopoles
opposite in polarity. According to Maxwell’s equation we cannot have magnetic monopoles,
though they are predicted by some modern theories, they have never been found. The trick is in
that by treating the North magnetic pole and South magnetic pole as separate magnetic charges
is to treat them like we do electric charges, the charge of a proton and the charge of an electron,
which can be convenient for making computations, but don’t exist that we know of. So we will
solve equation 56 for , and see what its units are so we can understand what it represents and
we will let m be the mass of a proton and q the charge of a proton. We get:
Equation 7.8
We can write these units as, by taking Coulombs (C) equal to
Equation 7.9
This is units of force per current density, which makes sense because a flowing current creates a
force. We can also write it:
Equation 7.10
Which is energy per magnetic field strength in that the SI units of magnetic field strength is
amps per meter. This tells us:
Equation 7.11
Thus we will use the energy as ionization of hydrogen, the energy to remove its electron and
make it a proton:
We have
, , ,
r
c
=
(
q
2
μ
2
2GM
m
2
)
1/3
μ
μ
r
3
GM
m
2
p
q
2
p
= μ =
m
3
kg
C s
a m p secon d s
kg
m
s
2
m
2
a m ps
kg
m
2
s
2
m
a m ps
μ =
En erg y
Magn et icFiel d Strength
H H
+
+ e
= 1proton = 2.18E 18 J
q
p
= 1.6E 19C
m
p
= 1.67E 27kg
G = 6.67408E 11N
m
2
kg
2
M
= 1.989E 30kg
! of 33 66
We want to look at the Sun as having a current flowing around its equator in a loop with its
radius
We find for the dipole field of the Sun at 50 Gauss=5E-3T, which is about 100 times stronger
than the Earth magnetic field, that this is a current I=5.5362E12 amperes driving the solar
magnetic dipole. This gives us that since the Earth orbit (1AU=1.495979E11m):
Equation 7.12
Equation 7.13
Or,…
Equation 7.14
The radius of a proton is 0.833E-15m. We have that r is:
Equation 7.15
R
= 6.957E8m
I
1AU
=
5.5362E12a mps
1.496E11m
= 37.0a m ps /m
μ =
Ioni z at ion Energ y
Magn et icFiel d Strength
=
2.18E 18J
37.0A m ps /m
= 5.892E 20
J m
A
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
r
r
p
= 5.92 6Pr oton Ra dii = carbon
! of 34 66
Our six-fold symmetry unfolding. This is again the carbon the core element of life. We see it
provided for by the Sun’s magnetic field.
!
! of 35 66
Let us write the computation as one equation, and verify it. We have
Where
=Ionization energy of Hydrogen
=3.47E-39 (correct)
Equation 7.16
(0.000034574)(3.47E-39)=1.19972E-43
1.19972E-43^(1/3)
=4.932E-15
So as you can see equation 7.15 is correct. It says that carbon, the basis of life is in the ratio of
the solar magnetic field and the solar gravitational field.!
r =
(
q
2
p
μ
2
2GM
m
2
p
)
1/3
= 4.9324E 15m
I
=
2B
R
μ
0
I
1AU
=
5.5362E12a m ps
1.496E11m
= 37.0 a m ps /m
1AU = r
e
μ =
Ioni z at ion Energ y
Magn et icFiel d Strength
=
2.18E 18J
37.0A m ps /m
= 5.892E 20
J m
A
IE
H
I
1AU
=
2B
R
μ
0
1
r
e
μ
2
= (5.982E 20)
2
= 3.47E 39
μ
2
=
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
=
(2.18E 18)
2
(12.56637E 7)
2
(1.496E11)
2
4(5E 3)
2
(6.957E8)
2
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
6r
p
=
(1.6E 19C )
2
2(6.67408E 11N
m
2
kg
2
)(1.989E 30kg)(1.67E 27)
2
(2.18E 18J )
2
(12.56637E 7T m /A)
2
(1.496E11m)
2
4(5E 3T )
2
(6.957E8m)
2
1/3
! of 36 66
Thus we have a theoretical value for the radius of the proton:
Equation 7.17
And a theoretical value for its charge
Equation 7.18
Equation 7.19
And we have the radius of a proton in terms of the solar magnetic field at Earth
Equation 7.20
All of this based on the idea that the basis of their structure is in six-fold unfolding. Given our
constant k
Equation 7.21
And making the approximation we can with these equations eliminate in equation
5.21 using equation 5.19 in which we can eliminate k with equation 2.22.
Equation 7.22
This has an accuracy of close to 88% because in equation 7.19 the charge of six protons is
predicted by the theory to be 5.72 protons, the rest of the equations are much more accurate, but
we seek to rectify that. We write this equation so we can have an equation that defines the solar
magnetic field by solving for .
To address the accuracy of the equation for the charge of a proton, equation 7.19, we ask what is
the culprit. We suggest it is . So we solve the equation for that to see by how much it is
off. It is:
Equation 7.23
We see it should be 7.79573E-11 and is actually 7.885E-11. The value we are using is 98.86785%
accurate, but we want to do better.
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71proton s 6proton s
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
1 + α = 1
q
p
31850496r
2
p
=
α
4
k
e
c
4
h
π
3
G
1
(N
A
𝔼)
2
h 4π
Gc
(IE
H
)
2
μ
2
0
r
2
e
GM
m
4
p
B
2
R
2
B
(α
4
)/36
α
4
36
= q
2
k
e
c
k
2
Gc
h 4π r
2
p
! of 37 66
We now eliminate on the left in equation 7.22 with equation 7.17 to find the B field of the Sun
as described by the Earth orbit as the ground state:
Equation 7.24
Where N=31850496 is a perfect integer, that is has no values after the decimal. Equation 8 gives
a magnetic field strength of 4.73E-3 Teslas. Concerning equation 7.24
We want it to be more accurate. To do that we have to substitute for , the fraction
which is 39/5 and we have to formulate a theory for why this would be, though the discrepancy
could be in as its experimental value has the largest errors. Other possibilities are the
is right and that as well this is due to the radius of a proton having large errors, or even that it is
supposed to be the factor 8, which would be good to consider because 8-fold symmetry is very
dynamic, in particular in its role with beryllium 8 being a precursor to carbon in nuclear
synthesis by stars.
Because we are looking at 6 protons we had , and because of six-fold symmetry we had
and because the charge we had because of the 1/3 root on the right. So taking
and leaving the that came from substituting for k, and the as
well, and leaving the factors 2 and 4 because they describe the physical dynamics of the equation
we have:
Equation 7.25
Where
Or we can factor out all the numbers on the right the 2 and 4 in equation 7.25 and write
Equation 7.26
In which case N=31850496/4=7962624. We see the solar magnetic field is determined by the
radius of a hydrogen atom, its ionization energy, and the solar gravitational field, with the earth
orbit as the ground state.!
r
2
p
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h 4π
Gc
α
4
36
= q
2
k
e
c
k
2
Gc
h 4π r
2
p
(α
4
)/36
7
4
5
r
p
(α
4
)/36
6
2
α
2
/6
q = 6q
p
6
3
N = 6
7
= 279936
3
2
/4
2
8 = 2
3
B
2
=
2
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
2GM
m
2
p
4R
2
3
2
4
2
α
4
c
3
2
3
k
e
h 4π
Gc
N = 6
7
= 279936
B
2
=
1
NR
2
H
(IE
H
)
2
μ
2
0
r
2
e
GM
m
2
p
R
2
α
4
c
3
k
e
h π
Gc
r
e
! of 38 66
Protons and neutrons pack in atoms in a way that can be seen from their electron clouds. Thus
since we can say the electron cloud of helium is spherical the nucleus packing is spherical. But
since the protons and and neutrons in the nucleus are moving (But only as far as their nuclear
wall) the packing is ordered but liquid.
I believe the reason for my finds of six-fold symmetry in both the radius of a proton and it its
charge can be explained by Buckminster Fuller’s vector equilibrium. It is the most
transformable straight-line geometry if you attach sticks with flexible corners, what Fuller calls a
flex-corner. See the following illustration.
Which means it zero-frequency for omni directional closest packing of spheres is 12 spheres,
which is carbon (six protons and six neutrons) which is the basis of life as we know it. Its
diameter is then six proton radii which we found provided for by the solar magnetic field. The
frequency is the number on any symmetrically concentric shell or layer and is given by
The more protons an element has the more neutrons it needs so the strong nuclear force in the
nucleus can overcome the mutual repulsion of protons, thus holding the atoms together.
10F
2
+ 2 = 10(1)
2
+ 2 = 12 = carbon
! of 39 66
Thus, to go over that of the solar magnetic field again, proposed by Hannes Alfven (1942), at the
time there was no known mechanism for it, but he suggested the relative velocity between a
neutral gas and a plasma has a critical velocity at which the gas starts to ionize and that the
atoms or molecules will not exceed this velocity until the gas becomes almost fully ionized. The
additional energy put into the system goes into ionizing the gas instead of the velocity of the
atoms, and is roughly independent of pressure and magnetic field. Critical ionization velocity
has been recognized in the laboratory for some time. It is given by equating the kinetic energy of
the atoms to the ionization potential:
Alfven found:
Gas cloud enters Solar System
A neutral atom falls into Sun due to gravity
Motion is random, collisions happen
Temperature rises
At a distance from the Sun gas will ionize
m=atomic weight
He found for a gas cloud with average voltage 12 volts. An average atomic weight of 7, the is at
Jupiter.
That is atoms fall in towards the Sun and ionize at which point the solar magnetic field pushes
them out to Jupiter orbit where a halo forms from which planets can form.
1
2
mv
2
= eV
ion
r
i
= G
Mm
eV
ion
r
i
= G
Mm
eV
ion
= 13.5E10
m
V
ion
cm
V
ion
= volts
r
i
! of 40 66
Thus if we are to suggest, as we have, that solar magnetic field provides for life, whose skeletons
are the hydrocarbons, that this has something to due with the solar wind which is plasma and
we can consider it a plasma that can move in neutral gas thus being subject to Alfven’s critical
velocity. The plasma is mostly electrons, protons, alpha particles kinetic energies between 0.5
and 10 keV. Their are trace amount sof heavy ions of C, N, O, Ne Mg, Is, S, and Fe. We note the
C, N, O are the most abundant elements in life chemistry. The solar wind can reach velocities of
25o000-750000 m/s. Remember the critical velocity for hydrogen is 50,900 m/s.
! of 41 66
8.0 Conclusion
Thus in conclusion we gave carbon as one second determines the radius of a proton
Where one second is
That this is nearly perfectly one second makes the second dynamic because carbon is
And hydrogen is
Which is dynamic because we have integer seconds in terms hydrocarbons the skeletons of life
being the upper and lower limits for integers that produce protons in elements. That is one
second is the lower integer limit that determines whole protons and is carbon the core element
of life and anything greater than six seconds is fractional protons and is hydrogen. Interestingly
the second we find is described in terms of the constants and atomic dimensions and in terms of
something on a much larger scale, the Earth/Moon/Sun orbital parameters and that these in
turn determine the radius of a proton:
If multiplied by one second ( ) gives
Where is
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352 secon d s
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton 6secon d s = hydr ogen(H )
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352 secon d s
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t
1
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
t
1
! of 42 66
Giving
We see our constant k, determines the radius of a proton on the microscale and the Earth orbit
in terms of the six-fold
Where is the Earth orbital velocity. We see that k is
Where is an equation of state for the periodic table of the elements and is equal to
Avogadro’s number, making it a Natural number, not just an arbitrary large number for dealing
with large amounts of atoms. We see k is determined by our idea of an intermediary mass
between protons and the stars using the Chandrasekhar limit for white dwarf stars. We see we
can use our constant k to determine the charge of a proton in terms of our six-fold:
And that we can do the same with the solar magnetic field:
Which is to suggest the solar magnetic field perhaps balances with its gravity to make six proton
radii functional at Earth orbit. We see this in terms of Buckminster Fuller’s vector equilibrium:
t =
R
H
2α
2
m
p
h 4π
Gc
r
p
= 8.26935E 16m 0.827f m
k v
e
= 6
v
e
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
N
A
𝔼
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71proton s 6proton s
6r
p
=
(
q
2
p
2GM
m
2
p
(IE
H
)
2
μ
2
0
r
2
e
4B
2
R
2
)
1/3
! of 43 66
9.0 The Radius of a Proton Revisited
In section 3.0 that starts on page 12, we noted to determine the radius of a proton we needed
an equation for the unit of one second in terms of the Natural constants. We found the radius
of a proton was"
If we multiplied through by one second. We are guessing one second is a natural unit because we
found it is in the orbital mechanics of the Earth and Moon and indeed we measure time by the
periods of their motions. We found
If it is calculated with orbital velocities at perihelion and at aphelion
which gives a near perfect result for one second and still close in the average
orbital velocities, which have little variation because the orbits of these bodies are nearly circular
(have low eccentricities).
Our value for one second in terms of the Natural constants uses the radius of a hydrogen atom
as given by the Van der Waals radius, the hydrogen atom being the simplest element with just
one proton which is the basic building block of the periodic table of the elements. It makes sense
in our theory here to use the Van der Waals Radius, because our theory is holistic; it determines
its parts from the whole, and the whole from the parts. So the proton is a part of the hydrogen
atom, the most fundamental part, and indeed we can call a hydrogen atom a proton, which we
actually do in chemistry, we call it a positive hydrogen ion, because stripped of its electron the
hydrogen ion is simply a proton.
Our theory is that the microcosmos is related to the macrocosmos and is a function of it as much
as the macrocosmos is a function of the microcosmos. But there are many levels that constitute a
cosmos; the proton theory is nuclear chemistry or atomic physics, but the Van der Waals radius
is chemistry, chemistry; the elements made from the atomic particles. So these are two levels.
Another level up from the elements are the compounds, and from there we go to the planets,
which takes us to the stars, and from the stars we go to galaxies, from there to clusters of
galaxies, from here we can go to the universe and from there to the multiverse.
And indeed we are bridging the different layers of the cosmos, we are doing it with our constant
k. We have earth at the ground state for sixfold symmetry:
Where k is
And k determines the time of one second in
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
v
e
= 30,290m /s
v
m
= 966m /s
k v
e
= 6
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
! of 44 66
Which we use to determine the radius of a proton
And is used to determine the charge of a proton
If we look at how we found that Avogadro’s number describes not just atoms, elements, and
compounds, for which it was devised in chemistry, we see it may describe all the levels of
cosmos, as we found in equations 4.5 and 4.6:
We see we can refine our equation for the radius of a proton by looking at the factors in these
equations. We can write them
Equation 9.1.
=
=
t =
R
H
2α
2
m
p
h 4π
Gc
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12 secon d s
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
c
k
e
(
α
2
6
1
k
)
2
h 4π r
2
p
Gc
= q
2
q
q
p
=
9.1435E 19C
1.602E 19C
= 5.71proton s 6proton s
N
A
𝔼 =
R
H
16π α
2
m
2
p
3 2
h
G
K E
earth
K E
moon
1
Ear th Da y
6.02E 23 =
R
H
16π α
2
m
2
p
3 2
h
G
K E
earth
K E
moon
1
Ear th Da y
K E
moon
K E
earth
=
R
H
16π α
2
m
2
p
3 2
h
GN
A
1
Ear th Da y
R
H
16π α
2
m
p
= 2.67875E19
m
kg
3.7331E 20
kg
m
2.225E 7
atom
cm
! of 45 66
Where we have considered meters/kg to be the atoms in a straight line approximated very thin
then cubed to get grams per volume as a good estimate for a density of atoms in space that can
be used to study the amount of atoms in the interstellar medium to the same in a nebula to the
same in a star or planet per unit volume. We write out h and G:
We have
=
(
2.225E 7
3
atom
cm
)
3
= 1.10E 22
atom s
cm
3
= 0.018
g
cm
3
= 18
kg
m
3
h = J s = kg
·
m
s
2
m s =
kg m
2
s
G = N
m
2
kg
2
= kg
m
s
2
m
2
kg
2
=
m
3
s
2
kg
N
A
= 6.02E 23
atom s
m ole
h
G
= kg
m
2
s
s
2
kg
m
3
= kg
2
s
m
h
G
=
6.62607E 34
6.67408E 11
= 9.928E 24kg
2
s
m
h
G m
p
=
9.928E 24
1.67262E 27
= 5935.6kg
s
m
R
H
16π α
2
m
p
= 2.67875E19
m
kg
h
G m
p
=
9.928E 24
1.67262E 27
= 5935.6kg
s
m
R
H
16π α
2
m
p
h
G m
p
=
(
2.67875E19
m
kg
)(
5935.6kg
s
m
)
1.5899885E 23secon d s 1.6E 23secon d s
N
A
= 6.02E 23
atom s
m ole
N
A
16π α
2
m
p
R
H
G m
p
h
=
6.02E 23atom s
1.6E 23secon d
= 3.7625
atom s
secon d
3 2 = 4.24264
! of 46 66
EarthDay=(24)(60)(60)=86400 seconds
We have
Equation 9.2.
That is
Equation 9.3.
Equation 9.4.
Now if we compare 9.3 with our old equation for a second we have
We see we eliminate the 16 and move into the denominator. We see that
The two are close to the same and this brings up the question; do we want the time unit be the
one closest to an actual second. The second comes from the Earth rotation dividing it into 60
units we call minutes and that into 60 units we call seconds. Certainly we see the second is
Natural, but do we want to change it a little as our fundamental unit. In terms of what we are
doing here, that depends on the exact value of the radius of a proton, and experimentally it is the
least accurate measurement we have been able to do in the laboratory to date among all our
constants used here. It depends on exactly what that is, and scientists believe they are really
(3.7525)(4.24264) = 15.962933 16
atom s
secon d
16s
1
(86400s)(16s
1
) = 1,382,400a tom s
N
A
3 2π α
2
m
p
R
H
G m
p
h
= 1
atom
sec
3.7625
16
(4.24264) = 0.9976833 1
R
H
h
3 2π α
2
m
2
p
GN
A
= 1secon d
N
A
=
R
H
h
3 2π α
2
m
2
p
Gt
1
= 6E 23
t
1
= 1secon d
R
H
16π α
2
m
2
p
3 2
N
A
𝔼
h
G
= 1.12 secon d s
3 2
3 2
16
= 0.265165
1
3 2
= 0.23570
! of 47 66
honing in on it. It wavers a little depending on the method we use to measure it. But I think it is
closest to the one using just because
Thus equation 9.3 can be written
Equation 9.5.
Which fits in best with our theory of sixfold symmetry because is the diagonal of a unit
square and we then have that divided up into six equal parts.
We have said our theory is holistic and that we are describing the proton radius in terms of the
whole of which it is a part, namely, the radius of a hydrogen atom, more specifically the Van der
Waals radius, which is determined by hydrogen gas, or . It is done like this:
3 2
1
3 2
=
2
6
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
2
H
2
! of 48 66
Johannes Diderik van der Waals (1873) described more than just Ideal gases, which are gases
that behave according to kinetic-molecular theory, he described real gases which don’t. His
equation then, The Van der Waals equation, is a modification of the Ideal Gas Law which is:
Which is quite obvious. If you increase the temperature T, then the volume of the gas is going to
increase, and if it doesn’t then the pressure will, which is inversely proportional to volume.
However for a Real Gas, he assumed the particles are hard spheres, cannot be compressed
beyond a limit, and at close proximity to one another they interact and have a volume around
them that excludes one another, that is they have walls. He said
That is the volume of the real gas ( ) is equal to the volume of the ideal gas ( ) minus a
correction factor b. The volume of the particles is the number of particles ( ) times the volume
of one particle:
Thus there exists a sphere of radius 2r formed by two particles in contact where no other
particles can enter. It gives the correction factor
And the volume correction for n particles is
This is the volume correction to the Ideal Gas Law. The pressure correction says real gases
exhibit less pressure because their particles interact which is a net pulling by the bulk of
particles away from the container walls.
The reduction in pressure is proportional to by a factor a. We have for reduction of pressure
that
We substitute this into the Ideal Gas Law:
This can be written as a cubic
PV = n RT
V
R
= V
I
b
V
R
V
I
n
n
4
3
π r
3
b = (4)
4
3
π r
3
nb = 4n ×
4
3
π r
3
n
2
v
2
P
I
= P
R
+ a
n
2
V
2
(
P + a
n
2
V
2
)
(
V n b
)
= n RT
! of 49 66
Which allows one to compute the critical conditions of liquefaction and toderive an expression
of the principle corresponding states. In the cubic form we have as the solution three volumes
which can be used for computing the volume at and below critical temperatures.
Thus the Van Der Waals radius is estimated
For hydrogen experimentally. Therefore with
We have described the derivation of radius of a hydrogen atom from the Van Der Waals
equations that we use to get
Where the Van Der Waals equations are
So we can compute the radius of a proton from
Which needs to be multiplied through by one second to get
V
3
(
b +
RT
P
)
V
2
+
a
p
V
a b
P
= 0
(
π +
3
φ
2
)
(3φ 1) = 8τ : π =
P
P
c
, φ =
V
V
c
, τ =
T
T
c
4
3
π r
3
w
=
b
N
A
r
3
w
=
3
4π
b
N
A
b = 26.61
cm
3
m ol
N
A
= 6.02E 23
r
w
= 1.0967E 8cm = 1.0967E 10m
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
(
P + a
n
2
V
2
)
(
V n b
)
= n RT
V
3
(
b +
RT
P
)
V
2
+
a
p
V
a b
P
= 0
4
3
π r
3
w
=
b
N
A
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
16
m = 0.829f m
! of 50 66
But in determining the radius of a proton from
We need to determine k in terms of the Chandrasekhar limit where k is
Where
Where we multiply by 1.44, the Chandrasekhar limit to get 2.8634E30kg giving us our
intermediate mass using the Chandrasekhar limit as the upper limit for a mass, and
The mass of a proton for the lower limit. To have our equation k
We needed to use the Chandrasekhar limit
So we address now how that limit is derived…
r
p
=
1
m
p
hc
2π
3
G
R
H
N
A
𝔼
r
p
=
18
3
α
2
m
p
Gc
4πh
t
1
= 8.288587 × 10
16
= 0.829f m
t =
R
H
2α
2
m
p
h 4π
Gc
t
ck
=
3 2
16
6.626E 34
6.674E 11
18769
(1.6726E 27)
2
1.2E 10
6.02E 23
1
π
= 1.12 secon d s
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
1 + α
1
N
A
𝔼
1
k
=
3
4
1
m
p
c
3
h
8π
3
G
1
N
A
𝔼
k =
1
m
2
i
h
1 + α
G
N
A
𝔼
m
i
= (2.8634E 30)(1.67262E 27) = 69.205kg
M
= 1.98847E 30kg
M
m
i
m
p
= 1.67262E 27kg
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) N
A
𝔼
M 0.77
c
3
3
G
3
N
m
4
p
= 1.41
! of 51 66
The pressure of the outer shell of star balances with the outward pressure in the core of
the star (thermal pressure). Pressure is force per unit surface area thus…
is the mass of the core pulling in the mass of the shell and is the radius of the
core. The surface area of the star is that of a sphere, . We have
The thermal pressure countering the gravity is given by the ideal gas law PV=nRT (pressure
times volume of a gas such as hydrogen , which is all protons , is proportional to temperature.
The number of protons in the core is . We have
Where is the Boltzmann constant ( ). Since we must have
if the star is not to implode or explode
And we have the estimate for the temperature of the core of a star.
Fusion would not occur at the low temperature of a star like the Sun in that there would not be
enough energy for collisions, unless the potential Coulomb barrier can be overcome by quantum
mechanical tunneling. The collisions are given by the kinetic energy of the particles
. We have
The velocity v yields the minimum distance between protons as the De Broglie wavelength
P
gravit y
P =
F
A
F = m a = PA
P =
m a
A
m a = G
M
shell
M
core
r
2
core
M
core
M
shell
r
core
A = 4π r
2
core
P
gravit y
= G
M
shell
M
core
4π r
4
core
m
p
N
p
M
core
m
p
P
ther mal
=
M
core
m
p
1
4
3
π r
3
core
k
B
T
core
k
B
1.380649E 23J K
1
P
gravit y
= P
ther mal
k
B
T
core
=
1
3
GM
shell
m
p
r
core
1
2
m
2
p
v
e
2
4πϵ
0
r
min
=
1
2
m
p
v
2
! of 52 66
Since the velocity is the root mean square velocity of the protons…
We have the temperature of the star is
This is another estimate. Since the mass of a star is its volume times its density
But for a star density varies with radius
If we take the derivative of both sides of the equation we have one of the equations of stellar
structure:
1.
The so-called conservation of mass equation. The force on the shell of the star is given by the
mass of the shell
Again for there to be balance gravitation pressure equals thermal pressure:
2.
Another equation of the equations of stellar structure. The so-called equation of hydrostatic
equilibrium. This can be written
If the star is an ideal gas the density of the star varies as where for a
monatomic gas and then
λ =
h
m
2
p
v
v
rms
=
3k
B
T
mp
T
min
=
(
e
2
4πϵ
0
)
2
m
p
3π
2
h
2
k
B
m =
4
3
π r
2
ρ
4π
r
0
r
2
ρ(r)dr
dm(r)
dr
= 4π r
2
ρ(r)
F
g
= G
m(r)4π r
2
ρ(r)
r
2
dr
dρ(r)
dr
= G
m(r)ρ(r)
r
2
d
dr
(
r
2
ρ(r)
dP(r)
dr
)
= 4π Gr
2
ρr
PV
γ
= consta nt
γ = 5/3
! of 53 66
In stellar dynamics we write
So that
The abundance of hydrogen and helium in the universe are approximately 75% and 24%,
respectively. Thus for every 4He2+ there are 12H+ and 2+12 free electrons. We have
Ionized hydrogen and helium have and for the Sun because of high metal
content. Finally stars can be approximated as blackbody radiators (purely radiate) and as such
pressure is given in terms of temperature (Temperature is proportional to radiation energy):
There are three kinds of pressures that can be generated by a star: gas pressure, radiation
pressure, or degeneracy pressure.
A type of star that is stable, that is prevented from collapse by degeneracy pressure, is a so-called
white dwarf star. They are the remnant of giant stars that have depleted the their fusion fuel and
thereby collapsed under gravity but are kept from collapsing into black holes by thermal
pressure due to motion of the particles alone. Interestingly, they still shine almost as bright as a
star on the main sequence even though they are not doing fusion. It was the Indian physicist
Chandrasekhar who found the limit in mass for which a white dwarf will not have its gravity
overcome the degeneracy pressure and collapse. The non-relativistic equation is:
There are many resources available that derive this and you can find it in any textbook on
astrophysics in the chapters dealing with stellar physics, and I will leave the treatment of the
derivation to those works.
Returning to our equation for one second in terms of the Earth and Moon orbital kinetic
energies:
P
1
V
γ
ρ
5/3
N
V
=
ρ
μm
p
P
gas
=
ρ
μm
p
k
B
T
4 + 12
1 + 12 + 14
= 0.59
μ = 0.59
μ = 0.62
P
rad
=
4
3
σ
c
T
4
M 0.77
c
3
h
3
G
3
N
m
4
p
= 1.41
K E
moon
K E
earth
(Ear th Da y) = 1.08secon d s
! of 54 66
We notice it would be exactly one second if the Earth day were shorter. And, indeed it was a long
time ago.
For our equation Earth day needs to be shorter. A long time ago it was; the Earth loses energy to
the moon. The days become longer by 0.0067 hours per million years. Our Equation
Is actually 1.2 seconds for KE of moon and KE or earth calculated with their average orbital
velocities. We will compute how long ago the Earth day was what it was needed to turn that 1.2
seconds into one second, and then how long ago the Earth day was what it was needed to turn
the 1.08 seconds into one second.
We have
24-20=0.0067t
t=597 million years
This was when the earth went through a dramatic change and there was a big explosion of life
(The Cambrian). This is when the Moon predicted the second as exactly 1. The dinosaurs went
extinct 65 million years ago giving small mammals a chance to evolve paving the way for
humans.
24-x=0.0067t
x=23.5645 hours
We say 20 hours + 3 hours is 0 hours + 3 hours since 20 hours is the zero of our cosmic
calendar:
What is the next term?
20+3+0.57735+0.4714=24hours
Which bring us to today. Now the calculation for the 1.08 seconds that works with KE of moon
and KE earth calculated using aphelions and perihelions.
K E
moon
K E
earth
(Ear th Da y) 1secon d
24h ours
1.2
= 20hours
3cos(0
) +
2
3
cos(30
) = din osaur ex t inct ion =
20hrs + 3hrs +
3
3
hrs +
2
3
=
24h ours
1.08
= 22.222h ours
! of 55 66
24-22.222=0.0067t
t=265.373 million years
This is the Permian period which was from 299 to 251 million years ago, it was at the end of the
Paleozoic Era which was followed by the Mesozoic Era. The distinction between the Paleozoic
and the Mesozoic is made at the end of the Permian to mark the largest mass extinction
recorded in Earth’s history.
Thus, to conclude, we write
And focus for a bit on . It takes the square which has diagonal to side is and divides that
up with regular hexagon which is six radii and six sides equation one another reconciling Euler’s
number e with golden ratio characteristic of fivefold symmetry, the regular pentagon. That is
Equation 9.6.
Now we show what this is geometrically and explore how we arrived at it in the following
pages…
2
6
R
H
h
π α
2
m
2
p
GN
A
= 1secon d
2
6
2
Φ
2
6
=
1
Φ(1 + Φ)
=
e
Φ
! of 56 66
! of 57 66
! of 58 66
! of 59 66
! of 60 66
! of 61 66
! of 62 66
!
! of 63 66
Appendix 1
We suggested the second was a natural unit, and that, if it was, should be in the orbital
mechanics of the earth because the second comes from the calendar, which is based on the
orbital period of the year (1 year, 365.25 days) and the orbital period of the moon, and the
rotation of the Earth. We found that it was as the following:
Let’s show that…
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.!
KE
moon
KE
earth
(EarthDay) 1second
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
! of 64 66
Appendix 2
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:!
1
α
2
m
p
h4π r
2
p
Gc
! of 65 66
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}!
! of 66 66
The Author
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